Standard Deviation Calculation – How to Calculate Standard Deviation for 1/x (Changed Unit)

meanstandard deviation

I have a report of a machine in which it is documented how many parts it has produced each day in the last year. From this I could calculate the mean with its standard deviation in [parts/day].
However, I need the mean and the standard deviation in [days/part], so mathematically "1/x".

Normally, I would first convert all daily values into the right unit, so [days/part] and then simply calculate the mean value and standard deviation.

However, this is not possible for various reasons in the further procedure of my work and so I first have to calculate the mean value and the standard deviation in [parts/day] and then convert them into the needed unit [days/part].

And here is my problem, I can convert the mean with "1/x", but that's not possible with the standard deviation.

My question now would be if there is a mathematically correct way to calculate the standard deviation in this case.

In the literature I unfortunately don't find any information about my topic/problem.

From my point of view, the only thing I can do is either estimate the standard deviation or maybe use the coefficient of variation of [parts/day]. Is that correct?

I hope, I'm not in the wrong place with my problem and that it is understandable. Any tips are welcome.

Best Answer

As the examples on https://math.stackexchange.com/questions/248472/expectation-on-1-x show, you cannot compute the mean of $1/x$ as $1/\mbox{mean}(x)$. Actually it is not possible to compute the mean of $1/x$ from merely konwing the mean and standard deviation (sd) of $x$. A mathematician would stop at this point ;-) As it is a real problem that needs a real solution, an engineer would nevertheless look for a workaround.

Although mean and sd cannot be converted to reciprocal values without knowing the entire distribution of $x$, the situation is different for the median $M$ and the interquartile range $IQR=x_{.75}-x_{.25}$. It is

$$ M\left(\frac{1}{x}\right) = \frac{1}{M(x)} \quad\mbox{ and }\quad IQR\left(\frac{1}{x}\right) = \frac{1}{x_{.25}} - \frac{1}{x_{.75}} = \frac{IQR(x)}{x_{.25}\cdot x_{.75}} $$

Apparently, your simulation software makes the assumption that the parameter can be merely described by mean and sd, and therefore most likely makes the assumption that it is normally distributed. For a normal distribution, mean and median are equal and $IQR \approx 1.34896 \sigma$, and you can approximate the standard deviation from the interquartile range, and use $1/M(x)$ as the mean.

This is only a very crude approximation and the distribution of $1/x$ is unlikely to actually be normal (or even symmetric), but I would expect it to become increasingly better when the coefficient of variation ($\sigma/\mu$) of $x$ is smaller.

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Solved – Average and standard deviation of timestamps (time wraps around at midnight)

Let's use the simplification you suggest: only use the data from positive readings and disregard the value of the reading, so we are left with a single set of circular data. You can use the circular dispersion, as whuber suggested, possibly multiplied by some constant to determine how much of the data should be seen as an outlier. A good text that is slightly easier to understand than the Wikipedia page would be Statistical Analysis of Circular data, by N.I. Fisher (1995).

I'll give some more straightforward formulas than the Wiki page, and give some sample code.

The dispersion can be calculated as (due to Fisher (p.32-34)):

Denote the data by $\boldsymbol\theta = \{\theta_1, \dots, \theta_n\}.$ An estimate of the mean $\bar\theta$ can be calculated with
- $S=\sum_{i=1}^{n}\sin(\theta_i)$,
- $C=\sum_{i=1}^{n}\cos(\theta_i)$,
- $\hat\mu = \text{atan2}(S, C)$. (See http://en.wikipedia.org/wiki/Atan2)
Calculate $\bar{R} = \frac{\sqrt{S^2 + C^2}}{n}$.
Calculate the dispersion as suggested by whuber. I'm not sure why, but Wikipedia's definition seems to differ slightly from Fisher. I will use Fisher's:
- $\hat\delta = \frac{1 - \left[ (1/n) \sum_{i=1}^{n} \cos 2 (\theta_i - \hat\mu) \right]}{2\bar{R}^2}.$
Then, choose some constant $c$ (1 is probably fine, but you may fine-tune). Then, the interval is given by
- $ \left[\hat\mu - c \hat\delta, \hat\mu + c \hat\delta \right]$.

I know you want to avoid R, but just to show how to calculate this in code, here is some basic R code anyway, which also generates a plot:

n  <- 200
th <- runif(n, 0.5 * pi, 1.5 * pi)

plot(cos(th), sin(th), xlim=c(-1, 1), ylim = c(-1, 1))

S <- sum(sin(th))
C <- sum(cos(th))

mu_hat <- atan2(S, C)

R_bar  <- sqrt(S^2 + C^2) / n

delta_hat <- (1 - sum(cos(2 * (th-mu_hat)))/n) / (2 * R_bar^2)

constant  <- 0.8

CI <- mu_hat + c(-1, 1) * constant * delta_hat

lines(x = c(0, cos(CI[1])), 
      y = c(0, sin(CI[1])), col="green")

lines(x = c(0, cos(CI[2])), 
      y = c(0, sin(CI[2])), col="blue")

Cut-off values for c=.8 with a uniform half-circle from a sample of 200.

In a final note, it may still be better to use the additional information provided by the value of the reading, and not only the sign, because it may provide better estimates. However, the simplification of only using the sign makes the problem much more manageable. If anyone has a good solution that incorporates the readings, I would love to know!

Statistics – Unit of Mean and Standard Deviation

Mean and standard deviation have dimensions, the same ones from your data. Take your example: $$ \text{mean distance } d \; [\text{in}\; m] = \mathbb{E}( d \; [\text{in}\; m]) = \cfrac{1}{n} \sum\limits_{j = 1}^{n} d_j \; [\text{in} \; m] = \mu \; [\text{in}\; m] $$

$$ \begin{aligned} \text{variance of distance } d \; [\text{in}\; m] &= \mathbb{E}(\, (d \; [\text{in}\; m] - \mu \; [\text{in}\; m])^2\,) \\ &= \cfrac{1}{n} \sum\limits_{j = 1}^{n} (d_j - \mu \; [\text{in} \; m] )^2 = \sigma^2 \; [\text{in}\; m^2] \\ &\Rightarrow \text{standard deviation } = \sqrt{\sigma^2 \; [\text{in}\; m^2]} = \sigma \; [\text{in}\; m] \end{aligned}$$

(for simplicity I've assumed that all distances in your dataset follow a uniform probability)

Best Answer

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Solved – Average and standard deviation of timestamps (time wraps around at midnight)

Statistics – Unit of Mean and Standard Deviation

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